Compile-time polymorphism in Java - Step-by-Step Execution

Concept Flow - Compile-time polymorphism

Start

↓

Call method with arguments

↓

Compiler checks method signatures

↓

Select method matching argument types

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Execute selected method

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End

At compile time, Java decides which method to call based on the method name and argument types.

Execution Sample

Java

class Demo {
  void show(int a) { System.out.println("int: " + a); }
  void show(String a) { System.out.println("String: " + a); }
  public static void main(String[] args) {
    Demo d = new Demo();
    d.show(5);
    d.show("Hi");
  }
}

This code shows two methods named show with different argument types; Java chooses which to call based on the argument.

Execution Table

Step	Method Call	Argument Type	Method Selected	Output
1	d.show(5)	int	show(int a)	int: 5
2	d.show("Hi")	String	show(String a)	String: Hi
3	End of main	-	-	-

💡 All method calls resolved at compile time by matching argument types.

Variable Tracker

Variable	Start	After Step 1	After Step 2	Final
d	null	Demo object	Demo object	Demo object
a (int)	N/A	5	N/A	N/A
a (String)	N/A	N/A	"Hi"	N/A

Key Moments - 2 Insights

Why does Java choose which show method to call before running the program?

Because Java uses compile-time polymorphism, it decides which method to call by looking at the argument types during compilation, as shown in execution_table steps 1 and 2.

What happens if you call show with a type that doesn't match any method?

The compiler will give an error because it cannot find a matching method signature, so no method is selected.

Visual Quiz - 3 Questions

Look at the execution table, which method is called when d.show(5) is executed?

A show(int a)

B show(String a)

C show(double a)

D No method is called

💡 Hint

Check the first row of the execution_table where argument type is int.

At which step does the method show(String a) get called?

A Step 1

B Step 2

C Step 3

D Never

💡 Hint

Look at the second row of the execution_table for method selection.

If we add a method show(double a), what happens when calling d.show(5.0)?

A show(int a) is called

B show(String a) is called

C show(double a) is called

D Compilation error

💡 Hint

Compile-time polymorphism selects method matching argument type exactly.

Concept Snapshot

Compile-time polymorphism (method overloading) means multiple methods have the same name but different parameters.
Java decides which method to call during compilation by matching argument types.
This allows methods to behave differently based on input types.
If no matching method is found, compilation fails.
Example: show(int) and show(String) are two overloaded methods.

Full Transcript

Compile-time polymorphism in Java means the program decides which method to call before running, based on the method name and argument types. In the example, the class Demo has two methods named show: one takes an int and the other a String. When main calls d.show(5), Java chooses show(int a) because the argument is an int. When d.show("Hi") is called, Java chooses show(String a). This decision happens during compilation, so the program knows exactly which method to run. If you try to call show with a type that doesn't match any method, the compiler will give an error. This is called method overloading and helps write clear code that works with different data types.